Question

William invested $12,000 in a bank account that pays 9 percent simple interest. His friend invested the same amount at another bank that pays 8 percent interest compounded quarterly. These two functions, where t is time in years, represent the value of the investments: f(t) = 12(1.02)4t g(t) = 12(1.09)t The functions are graphed, and the point of intersection lies between 0.5 and 1.2. Based on the table, approximately how long will it be until both investments have the same value? Value of t f(t) = 12(1.02)4t g(t) = 12(1.09)t 0.5 12.48 6.54 0.6 12.58 7.84 0.7 12.68 9.16 0.8 12.79 10.46 0.9 12.89 11.87 1.0 12.99 13.08 1.1 13.09 14.39 1.2 13.20 15.70 A. 0.9 year B. 1.0 year C. 1.1 years D. 1.2 years

Answer 1

Answer: B) 1.0 year

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Explanation:

We have these two functions

• f(t) = 12(1.02)^(4t)
• g(t) = 12(1.09)t

which represent the amounts for his friend and William in that order. Strangely your teacher mentions William first, but then swaps the order when listing the exponential function as the first. This might be slightly confusing.

The table of values is shown below. We have t represent the number of years and t starts at 0.5. It increments by 0.1

The f(t) and g(t) columns represent the outputs for those mentioned values of t. For example, if t = 0.5 years (aka 6 months) then f(t) = 12.48 and that indicates his friend has 12,480 dollars in the account.

I've added a fourth column labeled |f - g| which represents the absolute value of the difference of the f and g columns. If f = g, then f-g = 0. The goal is to see if we get 0 in this column or try to get as close as possible. This occurs when we get 0.09 when t = 1.0

So we don't exactly get f(t) and g(t) perfectly equal, but they get very close when t = 1.0

It turns out that the more accurate solution is roughly t = 0.9925 which is close enough. I used a graphing calculator to find this approximate solution.

It takes about a year for the two accounts to have the same approximate amount of money.